Question

am.nevoie urgent de ajutor exercițiu 30 tot urgent pana la 21 va rog

Answer 1

$a)\ 2^{x}=2\\x=1\\b)\ 2^{x}=4=2^{2}\\x=2\\c)\ 3^x=9=3^{2}\\\\x=2\\d)\ \sqrt{\frac{8*9}{2}}=\sqrt{36}=6^{2}\\x=2$

4^x=√(1+3(1+4+4^2+...+4^51))

Notam 1+4+4^2+...+4^51 cu s

s=1+4+4^2+...+4^51 /*4

4s=4+4^2+...+4^52

Scadem din a doua pe prima

3s=4^52-1

s=(4^52-1)/3

4^x=√(1+3*(4^52-1)/3)=√1+4^52-1=√4^52=4^26 => x=26

f) 5^x=√(1+4+4*5+...4*5^2011)=√(1+4(1+5+5^2+...+5^2011))

Notam s=1+5+5^2+...+5^2011

s=1+5+5^2+...+5^2011 /*5

5s=5+5^2+...+5^2012

4s=5^2012-1 => s=(5^2012-1)/4

5^x=√(1+5^2012-1)=√5^2012=5^1006 => x=1006

Answer 2

Pozele conțin rezolvarea.